(Original post by und)
I mentioned that I took m->∞ when I was finding the first limit; is that not enough? Or why can't we just write m=n?

Ordinarily, you would care what m is doing in the first (inner) limit, because as you've pointed out, the result you've stated only holds for . However, this isn't a problem overall, since when you look at the second limit...

(Original post by shamika)
Ordinarily, you would care what m is doing in the first (inner) limit, because as you've pointed out, the result you've stated only holds for . However, this isn't a problem overall, since when you look at the second limit...

(Original post by und)
The only thing that I see in that limit is m->∞, but I know nothing about repeated limits so I don't know what the conditions are for using this fact in the first limit.

In this case, assume you can use that fact

I think the thing that's confusing you is that you might be thinking that one limit is applied first and then the other. They're both being applied at the same time, that is, the following doesn't happen:

Think of it as both occurring simultaneously. To be honest, I was envisaging the difficulty of trying to swap the limits around, but in the case x is rational, it's ok to do so (at least, the result is the same either way)

However, I think you've sufficiently understood to be able to credit yourself with the answer. Let me know if you want me to write something up, but it'll essentially be what I've already tried to explain

(Original post by bananarama2)
I've got it. It you plot lateral binary graph associated with the symbolic mathematical solutions, you can reduced the problem down to simple SNF's. And by using Dalek's contraction algorithm the answer is: the physical quantity of dimensions m to the ml to the lq to the qt
to the ttheta to the x has planck quantity root g minus m plus l minus 2q
plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q.

You know the answer to a set theory question is wrong when it contains the word "planck"

Firstly, we have to apply the limit as n becomes arbitrarily large. As the power increases, there are one of three possibilities:

, in which case it will approach zero.

, in which case it will remain at one.

, in which case it will oscillate between -1 and 1.

Next, look at the different possibilities of m with regards to x. If x is rational, then sufficiently large m will contain the denominator of x and a multiple of 2, making x a multiple of so that the overall thing will reach a limit of 1. However, if x is irrational then it is impossible for it to become an integer multiple of . Because of this, cosx will approach zero as n tends towards infinity as the magnitude of cosx will be less than one.

Firstly, we have to apply the limit as n becomes arbitrarily large. As the power increases, there are one of three possibilities:

, in which case it will approach zero.

, in which case it will remain at one.

, in which case it will oscillate between -1 and 1.

Next, look at the different possibilities of m with regards to x. If x is rational, then sufficiently large m will contain the denominator of x and a multiple of 2, making x a multiple of so that the overall thing will reach a limit of 1. However, if x is irrational then it is impossible for it to become an integer multiple of . Because of this, cosx will approach zero as n tends towards infinity as the magnitude of cosx will be less than one.

You don't need the third case because you're looking at the limit of (and hence you can simplify the rest of the proof) But yes, that's essentially it